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Monday, November 14, 2016

An exercise about Lagrange mean value theorem

Problem:


Solution:

This is a really bored after lunch hour, so I decided let's do a simple calculus exercise to wake my brain up.

The 11+u2 reminded me this has something to do with tan. So let's do this simple integration.

du1+u2=sec2θdθ1+tan2θ=dθ=θ=tan1u.

That's refresh my memory! 11+u2 is the derivative of tan1u.

Now we apply the Lagrange's mean value theorem to get:

tan1vtan1uvu=11+δ2.

Where u<δ<v.

tan1vtan1u=vu1+δ2.

Therefore we conclude (remember when you increase the denominator, the value decrease, and vice versa):

vu1+v2<tan1vtan1u<vu1+u2.

Last but not least, set u=1 and v=43 gives

4311+(43)2<tan143tan11<4311+12.

325<tan143tan11<16.

π4+325<tan143<π4+16.

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